codeforces 871A Maximum splitting

codeforces 871A Maximum splitting

​​ You are given several queries. In the i-th query you are given a single positive integer ni. You are to represent ni as a sum of maximum possible number of composite summands and print this maximum number, or print -1, if there are no such splittings.

An integer greater than 1 is composite, if it is not prime, i.e. if it has positive divisors not equal to 1 and the integer itself.

Input The first line contains single integer q (1 ≤ q ≤ 105) — the number of queries.

q lines follow. The (i + 1)-th line contains single integer ni (1 ≤ ni ≤ 109) — the i-th query.

Output For each query print the maximum possible number of summands in a valid splitting to composite summands, or -1, if there are no such splittings.

Examples Input 1 12 Output 3 Input 2 6 8 Output 1 2 Input 3 1 2 3 Output -1 -1 -1 Note 12 = 4 + 4 + 4 = 4 + 8 = 6 + 6 = 12, but the first splitting has the maximum possible number of summands.

8 = 4 + 4, 6 can’t be split into several composite summands.

1, 2, 3 are less than any composite number, so they do not have valid splittings.

分类讨论一下 按照/4的余数来讨论

…1的情况那么可以用一个9减去 然后剩下的一定是4的倍数

…2的情况可以减去6 然后就是4的倍数了

…3的情况减去6+9然后是4的倍数了之所以要凑4的倍数就是贪心的想法啊

4是最小的符合要求的数

#includeint n;int main(){// freopen("cf.in","r",stdin); scanf("%d",&n); for (int i=1;i<=n;++i){ int ans=0;int tmp=0;scanf("%d",&tmp); if (tmp%4==3) { ans+=2;if (tmp<15) { printf("-1\n");continue; }tmp-=15;ans+=tmp/4;printf("%d\n",ans);continue; } if (tmp%4==2){ ans+=1;if (tmp<6){ printf("-1\n");continue; } tmp-=6;ans+=tmp/4;printf("%d\n",ans);continue; } if (tmp%4==1){ ans+=1;if (tmp<9) { printf("-1\n");continue; } tmp-=9;ans+=tmp/4;printf("%d\n",ans);continue; } if (tmp%4==0) { printf("%d\n",tmp/4);continue; } printf("-1\n"); } return 0;}

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